How do you solve #log(x+7) - log(x-7)=1#?

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Answer 1 · 2016-07-16T07:21:09

#x=77/9#

As #log(x+7)−log(x−7)=1#, we have

#log((x+7)/(x−7))=1 = log 10#

#(x+7)/(x−7)=10# or

#x+7=10x-70# or

#10x-x=77# or

#9x=77# or

#x=77/9#